Applications Of Derivatives Question 283
Question: Given function $ f(x)=( \frac{e^{2x}-1}{e^{2x}+1} ) $ is
[Orissa JEE 2005]
Options:
A) Increasing
B) Decreasing
C) Even
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ f(x)=\frac{e^{2x}-1}{e^{2x}+1} $
therefore (i) $ f(-x)=\frac{{e^{-2x}}-1}{{e^{-2x}}+1}=\frac{1-e^{2x}}{1+e^{2x}} $
therefore $ f(x)=-\frac{e^{2x}-1}{e^{2x}+1}=-f(x) $
$ f(x) $ is an odd function.
Again $ f(x)=\frac{e^{2x}-1}{e^{2x}+1}\Rightarrow {f}’(x)=\frac{4e^{2x}}{{{(1+e^{2x})}^{2}}}>0,\forall ,n\in R $
therefore $ f(x) $ is an increasing function.
BETA
